fourth edition mechanics of materials - …yunus.hacettepe.edu.tr/~boray/1_introduction...

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

Introduction –John T. DeWolf

Lecture Notes:Concept of Stress

J. Walt OlerTexas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Contents

Concept of Stress Bearing Stress in ConnectionsReview of StaticsStructure Free-Body DiagramC F B d Di

Stress Analysis & Design ExampleRod & Boom Normal StressesPi Sh i SComponent Free-Body Diagram

Method of JointsStress Analysis

Pin Shearing StressesPin Bearing StressesStress in Two Force MembersStress Analysis

DesignAxial Loading: Normal Stress

Stress in Two Force MembersStress on an Oblique PlaneMaximum StressesAxial Loading: Normal Stress

Centric & Eccentric LoadingShearing Stress

Maximum StressesStress Under General LoadingsState of Stressg

Shearing Stress Examples Factor of Safety

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 1- 2


FourthEdition


Concept of Stress

• The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures.g

• Both the analysis and design of a given structure involve the determination of stresses andinvolve the determination of stresses and deformations. This chapter is devoted to the concept of stress.



FourthEdition


Review of Statics

• The structure is designed to gsupport a 30 kN load

• The structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supports

• Perform a static analysis to determine the internal force in

j pp

each structural member and the reaction forces at the supports



FourthEdition


Structure Free-Body Diagram

• Structure is detached from supports and the loads and reaction forces are indicated

( ) ( )( )m8.0kN30m6.00 −==∑ xC AM

• Conditions for static equilibrium:( ) ( )( )

0

kN40

+==

=

∑

∑

xxx

x

xC

CAF

A

0kN300

kN40

=−+==

−=−=

∑ yyy

xx

CAF

AC

• Ay and Cy can not be determined from

kN30=+ yy CA

y y

these equations



FourthEdition


Component Free-Body Diagram

• In addition to the complete structure, each component must satisfy the conditions for p ystatic equilibrium

( )800∑ AM• Consider a free-body diagram for the boom:

( )

0

m8.00

=

−==∑

y

yB

A

AM

substitute into the structure equilibrium

kN30=yC

substitute into the structure equilibrium equation

• Results:↑=←=→= kN30kN40kN40 yx CCA

Reaction forces are directed along boom and rod



FourthEdition


Method of Joints• The boom and rod are 2-force members, i.e.,

the members are subjected to only two forces which are applied at member ends

• For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions

• Joints must satisfy the conditions for static ilib i hi h b d i th

0=∑ BFr

equilibrium which may be expressed in the form of a force triangle:

3kN30

54==

∑

BCAB

B

FF


kN50kN40 == BCAB FF


FourthEdition


Stress AnalysisCan the structure safely support the 30 kN

load?• From a statics analysis• From a statics analysis

FAB = 40 kN (compression) FBC = 50 kN (tension)

• At any section through member BC, the internal force is 50 kN with a force intensity

BC ( )

MPa159N105026

3=

×==

AP

BCσ

internal force is 50 kN with a force intensity or stress of

dBC = 20 mm

• From the material properties for steel, the allowable stress is

m10314 26-×ABC

• Conclusion: the strength of member BC is d

MPa 165all =σallowable stress is


adequate


FourthEdition


Design• Design of new structures requires selection of

appropriate materials and component dimensions to meet performance requirementsto meet performance requirements

• For reasons based on cost, weight, availability, etc., the choice is made to construct the rod from ,aluminum (σall= 100 MPa). What is an appropriate choice for the rod diameter?

3m10500

Pa10100N1050

2

266

3×=

×

×=== −

σσ

d

PAAP

allall

( ) mm225m10522m1050044

4

226

××

=

−−

π

Ad

dA

( ) mm2.25m1052.2 =×===ππ

d

• An aluminum rod 26 mm or more in diameter is


An aluminum rod 26 mm or more in diameter is adequate


FourthEdition


Axial Loading: Normal Stress

• The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis.

• The force intensity on that section is defined as

AP

AF

aveA

=∆∆

=→∆

σσ0

lim

the normal stress.

• The normal stress at a particular point may not be equal to the average stress but the resultant of the

AAA ∆→∆ 0

q gstress distribution must satisfy

∫∫ ===A

ave dAdFAP σσA

• The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics


alone.


FourthEdition


Centric & Eccentric Loading• A uniform distribution of stress in a section

infers that the line of action for the resultant of the internal forces passes through the centroidthe internal forces passes through the centroid of the section.

A if di t ib ti f t i l• A uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading.

• If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a ymoment.

• The stress distributions in eccentrically loaded


The stress distributions in eccentrically loaded members cannot be uniform or symmetric.


FourthEdition


Shearing Stress• Forces P and P’ are applied transversely to the

member AB.

• Corresponding internal forces act in the plane of section C and are called shearing forces.

• The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.

P=τ

• The corresponding average shear stress is,

and is equal to the load P.

A=aveτ

• Shear stress distribution varies from zero at the member surfaces to maximum values that may bemember surfaces to maximum values that may be much larger than the average value.

• The shear stress distribution cannot be assumed to


The shear stress distribution cannot be assumed to be uniform.


FourthEdition


Shearing Stress Examples

Single Shear Double Shear

FP==aveτ FP

ave ==τ


AAave AA 2aveτ


FourthEdition


Bearing Stress in Connections

• Bolts, rivets, and pins create stresses on the points of contactstresses on the points of contact or bearing surfaces of the members they connect.

• The resultant of the force distribution on the surface is

l d it t th f

C di f

equal and opposite to the force exerted on the pin.

• Corresponding average force intensity is called the bearing stress,

dtP

AP

==bσ



FourthEdition


Stress Analysis & Design Example

• Would like to determine the t i th b dstresses in the members and

connections of the structure shown.

• From a statics analysis:FAB = 40 kN (compression)

M t id i

FAB 40 kN (compression) FBC = 50 kN (tension)

• Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection



FourthEdition


Rod & Boom Normal Stresses• The rod is in tension with an axial force of 50 kN.

• At the rod center the average normal stress in theAt the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is σBC = +159 MPa.

( )( ) 10300254020 26−A

• At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,

( )( )

MPa167m10300

1050

m10300mm25mm40mm20

26

3,

26

=×

×==

×=−=

−N

AP

A

endBCσ

• The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.

m10300×A

g

• The minimum area sections at the boom ends are unstressed since the boom is in compression.


p


FourthEdition


Pin Shearing Stresses

• The cross-sectional area for pins at A, B, and C,

262

2 m104912mm25 −×=⎟

⎠⎞

⎜⎝⎛== ππ rA

• The force on the pin at C is equal to the force exerted by the rod BC,

MPa102m10491N1050

26

3, =

×

×== −A

PaveCτ

o ce e e ted by t e od C,

• The pin at A is in double shear with a total force equal to the force exerted bytotal force equal to the force exerted by the boom AB,

MPa7.40kN2026 ===

PaveAτ


m10491 26,× −AaveA


FourthEdition


Pin Shearing Stresses

• Divide the pin at B into sections to determine the section with the largest shear force,

(largest) kN25

kN15

=

=

G

E

P

P

kN50=BCF

• Evaluate the corresponding average

MPa9.50m10491

kN2526, =

×== −A

PGaveBτ

shearing stress,

m10491×A



FourthEdition


Pin Bearing Stresses

• To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

kN40P( )( ) MPa3.53

mm25mm30kN40

===tdP

bσ

T d t i th b i t t A i th b k t• To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,

MPa032kN40===

Pbσ ( )( ) MPa0.32

mm25mm50tdbσ



FourthEdition


Stress in Two Force Members

• Axial forces on a two force b lt i l lmember result in only normal

stresses on a plane cut perpendicular to the member axis.

• Transverse forces on bolts and pins result in only shear stresses p yon the plane perpendicular to bolt or pin axis.

• Will show that either axial or transverse forces may produce both

l d h i hnormal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.



FourthEdition


Stress on an Oblique Plane• Pass a section through the member forming

an angle θ with the normal plane.

• From equilibrium conditions, the distributed forces (stresses) on the plane

t b i l t t th f P

• Resolve P into components normal and i l h bli i

must be equivalent to the force P.

Th l d h

θθ sincos PVPF ==tangential to the oblique section,

θθσ coscos 2PPF

• The average normal and shear stresses on the oblique plane are

θθθτ

θ

θ

σθ

cossinsin

cos

cos 00

PPV

AAA

===

===


θθ

θ

τθ

cossin

cos 00 AAA===


FourthEdition


Maximum Stresses

• Normal and shearing stresses on an oblique plane

θθτθσ cossincos0

2

0 AP

AP

==

plane

• The maximum normal stress occurs when the reference plane is perpendicular to the member axisaxis,

00

m =′= τσAP

• The maximum shear stress occurs for a plane at + 45o with respect to the axis,

PP στ ′===00 2

45cos45sinAP

AP

m



FourthEdition


Stress Under General Loadings• A member subjected to a general

combination of loads is cut into t t b l itwo segments by a plane passing through Q

F x∆

• The distribution of internal stress components may be defined as,

VV

AF

xxy

x

Ax

∆∆

∆∆

=→∆

lim0

σ

AV

AV z

Axz

y

Axy ∆

∆=

∆

∆=

→∆→∆limlim

00ττ

• For equilibrium, an equal and opposite internal force and stress distribution must be exerted on


the other segment of the member.


FourthEdition


State of Stress• Stress components are defined for the planes

cut parallel to the x, y and z axes. For equilibrium equal and opposite stresses areequilibrium, equal and opposite stresses are exerted on the hidden planes.

• The combination of forces generated by the g ystresses must satisfy the conditions for equilibrium:

0=== ∑∑∑ zyx FFF

0

0

=== ∑∑∑

∑∑∑

zyx

zyx

MMM

• Consider the moments about the z axis:( ) ( )

yxxy

yxxyz aAaAM

ττ

ττ

=

∆−∆==∑ 0• Consider the moments about the z axis:

• It follows that only 6 components of stress are

yy

zyyzzyyz ττττ == andsimilarly,


o ows a o y 6 co po e s o s ess a erequired to define the complete state of stress


FourthEdition


Factor of Safety

Structural members or machines must be designed such that the

Factor of safety considerations:• uncertainty in material propertiesmust be designed such that the

working stresses are less than the ultimate strength of the material.

• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses

stress ultimate

safety ofFactor

u ==

=

σFS

FS • number of loading cycles• types of failure• maintenance requirements and

stress allowableall==

σFS maintenance requirements and

deterioration effects• importance of member to integrity of

h l t twhole structure• risk to life and property• influence on machine function


fourth edition mechanics of materials - …yunus.hacettepe.edu.tr/~boray/1_introduction...

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